Re: [PATCH v2] tracing: Optimize event type allocation with IDA
From: Zheng Yejian <hidden>
Date: 2022-11-23 08:01:11
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On Tue, 22 Nov 2022 22:32:58 -0500 Steven Rostedt [off-list ref] wrote:
On Wed, 23 Nov 2022 11:18:06 +0800 Zheng Yejian [off-list ref] wrote:quoted
But Yujie Liu [off-list ref] reported a problem that highly reproducible after applying this patch: Link: https://lore.kernel.org/lkml/54f23c9c-97ae-e326-5873-bfa5d2c81f52@intel.com/ (local) So please DO NOT apply this patch before I find what happened about it.I know what the issue is. The current way of assigning types is to always increment. And not to reuse until it fills up. And even then, it looks for the next available number. I'm guessing the IDA will reuse a number as soon as it is freed. This
Yes, it is.
may also have uncovered a bug, as in reality, we must actually clear the tracing buffers every time a number is reused.
Thanks for your explanation! It seems almost the case, and with current way of assigning types, this problem maybe also happend when reuse type id, am I right? But instead of clear tracing buffers, would it be better to just mark that record invalid if we had a way of knowing that the format had changed?
What happens is that the type number is associated to a print format. That is, the raw data is tagged with the type. This type maps to how to parse the raw data. If you have a kprobe, it creates a new type number. If you free it, and create another one. With the IDA, it is likely to reassign the previously freed number to a new probe. To explain this better, let's look at the following scenario: echo 'p:foo val=$arg1:u64' > kprobe_events echo 1 > events/kprobes/foo/enable sleep 1 echo 0 > events/kprobes/foo/enable echo 'p:bar val=+0($arg1):string' > kprobe_events # foo kprobe is deleted and bar is created and # with IDA, bar has the same number for type as foo cat trace When you read the trace, it will see a binary blob representing an event and marked with a type. Although the event was foo, it will now map it to bar. And it will read foo's $arg1:u64 as bar's +0($arg1):string, and will crash. -- Steve